You ordered an oligonucleotide primer (GCGTGGATCCATGTTTGCGG) from a company name

Question

You ordered an oligonucleotide primer (GCGTGGATCCATGTTTGCGG) from a company named Invitrogen-Life Technologies. The company provides you with the oligonucleotide in a lyophilized (dried) powder form. They tell you that they have sent you a total of 63,000 pmol of purified oligonucleotide that weighs 606 ug. What volume of buffer would you resuspend the oligonucleotide powder in order to make a 1mM stock solution? Please show work

Hint: You are given more information than you need to answer his question. Also, think of the oligonucleotide as any reagent (e.g. NaCL of Tris buffer); don't get caught up with being a different kind of reagent than you have worked with before.

SHOW WORK PLEASE

Explanation / Answer

Ans. Given,

            Moles of oligonucleotide = 63000 picomol

= 63000 x 1012 mol                         ; [1 picomol = 10-12 mol]

                                                = 6.30 x 10-8 mol

Total mass of oligonucleotide = 606 µg

                                                = 6.06 x 10-4 g                                   ; [1 µg = 10-6 g]

To prepare 1.0 mM solution.

That is 10-3 moles of solute (oligonucleotides) in 1.0 L of solution.

#. So far, we have,

            Desired molarity of solution = 10-3 M

            Available number of moles = 6.30 x 10-8 mol

Let the volume by Y liters.

So,

            Molarity of solution = Moles of oligonucleotide / Volume (in liters) of solution

            Or, 10-3 M = (6.30 x 10-8 mol) / Y L

            Or, 10-3 mol/ L = (6.30 x 10-8 mol) / Y L                            ; [1 M = 1 mol/L]

            Or, Y = (6.30 x 10-8)/ 10-3 = 6.30 x 10-5

            Or, Y = 6.30 x 10-5

Hence, required volume = Y liter = 6.30 x 10-5 L

                                                = 63.0 µL                                           ; [1 L =106 µL]

Thus, required volume of solution = 63.0 µL

Note: It is NOT possible to prepare such solution.

Reason-

            I. Total volume of solution need to be 63.0 µL

            II. Mass of solute (oligonucleotide) = 606 µg. Assuming its density to be equal to that of water (1.0 g/mL), the volume of 606 µg = 606 µL. So, the volume of solute alone increases around 10 times the final volume of solution (630. 606 µL)

            III. Suggestion- Check if the required concentration of stock solution is 1mM (1 millimolar) or 1.0 µM (1.0 micromolar).

            IV. In case the stock concertation of 1.0 µM (1.0 micromolar) BUT NOT millimolar.

            Molarity of solution = Moles of oligonucleotide / Volume (in liters) of solution

            Or, 10-3 µM = (6.30 x 10-8 mol) / Y L

            Or, 10-6 mol/ L = (6.30 x 10-8 mol) / Y L                            ; [1 M = 1 mol/L]

            Or, Y = (6.30 x 10-8)/ 10-6 = 6.30 x 10-2

            Or, Y = 6.30 x 10-2

            Hence, required volume = Y liter = 6.30 x 10-2 L

                                    = 63.0 mL                                          ; [1 L = 103 mL]

= 63000 µL                                       ; [1 mL =103 µL]

Thus, required volume of solution = 63000.0 µL

            Volume of buffer required = final Volume of stock solution – volume of solute

                                                            = 63000.0 µL – 606 µL

                                                            = 62394 µL

Preparation of 1µM stock solution: Add 606 µL (= 606 µg) to 62394 µL buffer. The final solution is your desired stock solution.   

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.